NWhat is the Apparent Weight of a Bicyclist on a Hump?

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Homework Help Overview

The problem involves a bicyclist and their bicycle, with a combined mass of 62 kg, going over a hump with a radius of 6 m at a speed of 5 m/s. The focus is on determining the apparent weight at the top of the hump, which relates to the forces acting on the bicyclist in this scenario.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of centripetal force and its relation to apparent weight, questioning whether other forces, such as normal force, should be considered. There is exploration of the relationship between speed, gravity, and the forces acting on the bicyclist at the top of the hump.

Discussion Status

Some participants have provided guidance on how to approach the problem, suggesting that the normal force is the apparent weight being sought. There is an ongoing exploration of the forces involved, with participants checking their understanding and calculations.

Contextual Notes

Participants are navigating the definitions of apparent weight and normal force, with some uncertainty about how to incorporate all relevant forces into their calculations. The discussion reflects a mix of confidence and confusion regarding the problem setup and the necessary steps to find a solution.

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A bicycle of bicyclist of combined masss 62 kg go over a hump of radius 6m at a speed of 5m/s. what is their combined appararent weight at the top of the hump?

is it just...

Fc = MAc

Fc =M(V^2/r)

Fc = 62(25/6)

Fc = 290N
 
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You only took into account the centripetal force, there is another force acting on the system.

BTW "apparent weight" usually means that they are asking for the normal force (the reaction force that the hump applies).
 
Last edited:
grrr, I have no idea then.
 
Well think about it. If i stand on a hump and don't move then my weight is going to be w = mg but if i travel over the hump at some speed then the centripetal force will act in the opposite direction of gravity and make me lighter.
 
oh, then normal force is what i solved for. didn't I?
 
hmmm, ya, that's correct. Hold on let me check the problem again.
 
man, I have no idea how to solve this one..
 
you solved for the centripetal force

Sum of force = MA (a = 0 because no acceleration in y direction, or you can plug in a = v^2/r and keep left side as n - mg)

N - mg + Centripetal = 0

you already solved the centripetal force. Now plug in mg and solve for N to get "apparent weight" or normal force.
 
hmmm...thanks man. I appreciate it.
 
  • #10
anyways...

Fn - mg + Fc = 0

Fn = mg - Fc

Fn = 607.6-258.33

Fn is roughly 350
 

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